ZOJ 1655 FZU 1125 Transport Goods

迪杰斯特拉最短路径。

1.every city must wait till all the goods arrive, and then transport the arriving goods together with its own goods to the next city.  这个条件貌似可以忽略掉。

2.One city can only transport the goods to one city.  这个条件貌似也可以忽略掉，是一定的。

例如：A物品和B物品在不同的城市，他们都往X城市运输，然而X城市到首都的最少损失的路必定只有一条的，所以2条件可以忽略。至于1条件，等和不等好像也没什么区别...... = =！

综上所述：就是每个物品只管自己运输，算最短路，每条路的权值就是比例，算出每个城市到首都损耗最少的路即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = ;
double weight[maxn], ratio[maxn][maxn], e[maxn];
int s[maxn];
int main()
{
    int n, m, i, j, u, v, ii;
    double cc;
    while (~scanf("%d%d", &n, &m))
    {
        memset(s, , sizeof(s));
        memset(ratio, , sizeof(ratio));
        for (i = ; i <= n - ; i++) scanf("%lf", &weight[i]);
        for (i = ; i <= m; i++)
        {
            scanf("%d%d%lf", &u, &v, &cc);
            cc =  - cc;
            if (cc > ratio[u][v])
            {
                ratio[u][v] = cc;
                ratio[v][u] = cc;
            }
        }
        for (i = ; i <= n; i++) e[i] = ratio[n][i];
        e[n] = ; s[n] = ;
        for (ii = ; ii < n; ii++)
        {
            double  maxn = -;
            int flag = , x;
            for (i = ; i <= n; i++)
            {
                if (!s[i] && (maxn< || maxn<e[i]))
                {
                    flag = ;
                    maxn = e[i];
                    x = i;
                }
            }
            if (!flag) break;
            s[x] = ;
            for (i = ; i <= n; i++)
            if (!s[i] && ratio[x][i] !=  && e[x] * ratio[x][i] > e[i])
                e[i] = e[x] * ratio[x][i];
        }
        double anss = ;
        for (i = ; i < n; i++) anss = anss + e[i] * weight[i];
        printf("%.2f\n", anss);
    }
    return ;
}